(** * Logic: Logic in Coq *)
(* $Date: 2012-07-22 18:36:58 -0400 (Sun, 22 Jul 2012) $ *)
Require Export "Ch05_Prop".
(** Coq's built-in logic is extremely small: only [Inductive]
definitions, universal quantification ([forall]), and
implication ([->]) are primitive, while all the other familiar
logical connectives -- conjunction, disjunction, negation,
existential quantification, even equality -- can be defined using
just these. *)
(* ########################################################### *)
(** * Quantification and Implication *)
(** In fact, [->] and [forall] are the _same_ primitive! Coq's [->]
notation is actually just a shorthand for [forall]. The [forall]
notation is more general, because it allows us to _name_ the
hypothesis. *)
(** For example, consider this proposition: *)
Definition funny_prop1 :=
forall n, forall (E : beautiful n), beautiful (n+3).
(** If we had a proof term inhabiting this proposition, it would
be a function with two arguments: a number [n] and some evidence
that [n] is beautiful. But the name [E] for this evidence is not
used in the rest of the statement of [funny_prop1], so it's a bit
silly to bother making up a name. We could write it like this
instead, using the dummy identifier [_] in place of a real
name: *)
Definition funny_prop1' :=
forall n, forall (_ : beautiful n), beautiful (n+3).
(** Or, equivalently, we can write it in more familiar notation: *)
Definition funny_prop1'' :=
forall n, beautiful n -> beautiful (n+3).
(** This illustrates that "[P -> Q]" is just syntactic sugar for
"[forall (_:P), Q]". *)
(* ########################################################### *)
(** * Conjunction *)
(** The logical conjunction of propositions [P] and [Q] can be
represented using an [Inductive] definition with one
constructor. *)
Inductive and (P Q : Prop) : Prop :=
conj : P -> Q -> (and P Q).
(** Note that, like the definition of [ev] in the previous
chapter, this definition is parameterized; however, in this case,
the parameters are themselves propositions, rather than numbers. *)
(** The intuition behind this definition is simple: to
construct evidence for [and P Q], we must provide evidence
for [P] and evidence for [Q]. More precisely:
- [conj p q] can be taken as evidence for [and P Q] if [p]
is evidence for [P] and [q] is evidence for [Q]; and
- this is the _only_ way to give evidence for [and P Q] --
that is, if someone gives us evidence for [and P Q], we
know it must have the form [conj p q], where [p] is
evidence for [P] and [q] is evidence for [Q].
Since we'll be using conjunction a lot, let's introduce a more
familiar-looking infix notation for it. *)
Notation "P /\ Q" := (and P Q) : type_scope.
(** (The [type_scope] annotation tells Coq that this notation
will be appearing in propositions, not values.) *)
(** Consider the "type" of the constructor [conj]: *)
Check conj.
(* ===> forall P Q : Prop, P -> Q -> P /\ Q *)
(** Notice that it takes 4 inputs -- namely the propositions [P]
and [Q] and evidence for [P] and [Q] -- and returns as output the
evidence of [P /\ Q]. *)
(** Besides the elegance of building everything up from a tiny
foundation, what's nice about defining conjunction this way is
that we can prove statements involving conjunction using the
tactics that we already know. For example, if the goal statement
is a conjuction, we can prove it by applying the single
constructor [conj], which (as can be seen from the type of [conj])
solves the current goal and leaves the two parts of the
conjunction as subgoals to be proved separately. *)
Theorem and_example :
(beautiful 0) /\ (beautiful 3).
Proof.
apply conj.
(* Case "left". *) apply b_0.
(* Case "right". *) apply b_3. Qed.
(** Let's take a look at the proof object for the above theorem. *)
Print and_example.
(* ===> conj (beautiful 0) (beautiful 3) b_0 b_3
: beautiful 0 /\ beautiful 3 *)
(** Note that the proof is of the form
conj (beautiful 0) (beautiful 3)
(...pf of beautiful 3...) (...pf of beautiful 3...)
as you'd expect, given the type of [conj]. *)
(** Just for convenience, we can use the tactic [split] as a shorthand for
[apply conj]. *)
Theorem and_example' :
(ev 0) /\ (ev 4).
Proof.
split.
Case "left". apply ev_0.
Case "right". apply ev_SS. apply ev_SS. apply ev_0. Qed.
(** Conversely, the [inversion] tactic can be used to take a
conjunction hypothesis in the context, calculate what evidence
must have been used to build it, and add variables representing
this evidence to the proof context. *)
Theorem proj1 : forall P Q : Prop,
P /\ Q -> P.
Proof.
intros P Q H.
inversion H as [HP HQ].
apply HP. Qed.
(** **** Exercise: 1 star, optional (proj2) *)
(* EXPECTED *)
Theorem proj2 : forall P Q : Prop,
P /\ Q -> Q.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
Theorem and_commut : forall P Q : Prop,
P /\ Q -> Q /\ P.
Proof.
(* WORKED IN CLASS *)
intros P Q H.
inversion H as [HP HQ].
split.
(* Case "left". *) apply HQ.
(* Case "right".*) apply HP. Qed.
(** Once again, we have commented out the [Case] tactics to make the
proof object for this theorem easy to understand. Examining it
shows that all that is really happening is taking apart a record
containing evidence for [P] and [Q] and rebuilding it in the
opposite order: *)
Print and_commut.
(* ===>
and_commut =
fun (P Q : Prop) (H : P /\ Q) =>
let H0 := match H with
| conj HP HQ => conj Q P HQ HP
end
in H0
: forall P Q : Prop, P /\ Q -> Q /\ P *)
(** **** Exercise: 2 stars (and_assoc) *)
(* EXPECTED *)
(** In the following proof, notice how the _nested pattern_ in the
[inversion] breaks the hypothesis [H : P /\ (Q /\ R)] down into
[HP: P], [HQ : Q], and [HR : R]. Finish the proof from there: *)
Theorem and_assoc : forall P Q R : Prop,
P /\ (Q /\ R) -> (P /\ Q) /\ R.
Proof.
intros P Q R H.
inversion H as [HP [HQ HR]].
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, recommended (even__ev) *)
(* EXPECTED *)
(** Now we can prove the other direction of the equivalence of [even]
and [ev], which we left hanging in chapter [Prop]. Notice that the
left-hand conjunct here is the statement we are actually interested
in; the right-hand conjunct is needed in order to make the
induction hypothesis strong enough that we can carry out the
reasoning in the inductive step. (To see why this is needed, try
proving the left conjunct by itself and observe where things get
stuck.) *)
Theorem even__ev : forall n : nat,
(even n -> ev n) /\ (even (S n) -> ev (S n)).
Proof.
(* Hint: Use induction on [n]. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 2 stars, optional (conj_fact) *)
(** Construct a proof object demonstrating the following proposition. *)
Definition conj_fact : forall P Q R, P /\ Q -> Q /\ R -> P /\ R :=
(* FILL IN HERE *) admit.
(** [] *)
(* ###################################################### *)
(** ** Iff *)
(** The familiar logical "if and only if" is just the
conjunction of two implications. *)
Definition iff (P Q : Prop) := (P -> Q) /\ (Q -> P).
Notation "P <-> Q" := (iff P Q)
(at level 95, no associativity) : type_scope.
Theorem iff_implies : forall P Q : Prop,
(P <-> Q) -> P -> Q.
Proof.
intros P Q H.
inversion H as [HAB HBA]. apply HAB. Qed.
Theorem iff_sym : forall P Q : Prop,
(P <-> Q) -> (Q <-> P).
Proof.
(* WORKED IN CLASS *)
intros P Q H.
inversion H as [HAB HBA].
split.
Case "->". apply HBA.
Case "<-". apply HAB. Qed.
(** **** Exercise: 1 star, optional (iff_properties) *)
(* EXPECTED *)
(** Using the above proof that [<->] is symmetric ([iff_sym]) as
a guide, prove that it is also reflexive and transitive. *)
Theorem iff_refl : forall P : Prop,
P <-> P.
Proof.
(* FILL IN HERE *) Admitted.
Theorem iff_trans : forall P Q R : Prop,
(P <-> Q) -> (Q <-> R) -> (P <-> R).
Proof.
(* FILL IN HERE *) Admitted.
(** Hint: If you have an iff hypothesis in the context, you can use
[inversion] to break it into two separate implications. (Think
about why this works.) *)
(** [] *)
(** **** Exercise: 2 stars, optional (beautiful_iff_gorgeous) *)
(** We have seen that the families of propositions [beautiful] and
[gorgeous] actually characterize the same set of numbers.
Prove that [beautiful n <-> gorgeous n] for all [n]. Just for
fun, write your proof as an explicit proof object, rather than
using tactics. (_Hint_: if you make use of previously defined
theorems, you should only need a single line!) *)
Definition beautiful_iff_gorgeous :
forall n, beautiful n <-> gorgeous n :=
(* FILL IN HERE *) admit.
(** [] *)
(** Some of Coq's tactics treat [iff] statements specially, thus
avoiding the need for some low-level manipulation when reasoning
with them. In particular, [rewrite] can be used with [iff]
statements, not just equalities. *)
(* ############################################################ *)
(** * Disjunction *)
(** Disjunction ("logical or") can also be defined as an
inductive proposition. *)
Inductive or (P Q : Prop) : Prop :=
| or_introl : P -> or P Q
| or_intror : Q -> or P Q.
Notation "P \/ Q" := (or P Q) : type_scope.
(** Consider the "type" of the constructor [or_introl]: *)
Check or_introl.
(* ===> forall P Q : Prop, P -> P \/ Q *)
(** It takes 3 inputs, namely the propositions [P], [Q] and
evidence of [P], and returns, as output, the evidence of [P \/ Q].
Next, look at the type of [or_intror]: *)
Check or_intror.
(* ===> forall P Q : Prop, Q -> P \/ Q *)
(** It is like [or_introl] but it requires evidence of [Q]
instead of evidence of [P]. *)
(** Intuitively, there are two ways of giving evidence for [P \/ Q]:
- give evidence for [P] (and say that it is [P] you are giving
evidence for -- this is the function of the [or_introl]
constructor), or
- give evidence for [Q], tagged with the [or_intror]
constructor. *)
(** Since [P \/ Q] has two constructors, doing [inversion] on a
hypothesis of type [P \/ Q] yields two subgoals. *)
Theorem or_commut : forall P Q : Prop,
P \/ Q -> Q \/ P.
Proof.
intros P Q H.
inversion H as [HP | HQ].
Case "left". apply or_intror. apply HP.
Case "right". apply or_introl. apply HQ. Qed.
(** From here on, we'll use the shorthand tactics [left] and [right]
in place of [apply or_introl] and [apply or_intror]. *)
Theorem or_commut' : forall P Q : Prop,
P \/ Q -> Q \/ P.
Proof.
intros P Q H.
inversion H as [HP | HQ].
Case "left". right. apply HP.
Case "right". left. apply HQ. Qed.
(** **** Exercise: 2 stars, optional (or_commut'') *)
(** Try to write down an explicit proof object for [or_commut] (without
using [Print] to peek at the ones we already defined!). *)
(* FILL IN HERE *)
(** [] *)
Theorem or_distributes_over_and_1 : forall P Q R : Prop,
P \/ (Q /\ R) -> (P \/ Q) /\ (P \/ R).
Proof.
intros P Q R. intros H. inversion H as [HP | [HQ HR]].
Case "left". split.
SCase "left". left. apply HP.
SCase "right". left. apply HP.
Case "right". split.
SCase "left". right. apply HQ.
SCase "right". right. apply HR. Qed.
(** **** Exercise: 2 stars, recommended (or_distributes_over_and_2) *)
(* EXPECTED *)
Theorem or_distributes_over_and_2 : forall P Q R : Prop,
(P \/ Q) /\ (P \/ R) -> P \/ (Q /\ R).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (or_distributes_over_and) *)
Theorem or_distributes_over_and : forall P Q R : Prop,
P \/ (Q /\ R) <-> (P \/ Q) /\ (P \/ R).
Proof.
(* FILL IN HERE *) Admitted.
Theorem orb_true : forall b c,
orb b c = true -> b = true \/ c = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem orb_false : forall b c,
orb b c = false -> b = false /\ c = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################### *)
(** * Falsehood *)
(** Logical falsehood can be represented in Coq as an inductively
defined proposition with no constructors. *)
Inductive False : Prop := .
(** Intuition: [False] is a proposition for which there is no way
to give evidence. *)
(** **** Exercise: 1 star (False_ind_principle) *)
(* EXPECTED *)
(** Can you predict the induction principle for falsehood? *)
(* Check False_ind. *)
(** [] *)
(** Since [False] has no constructors, inverting an assumption
of type [False] always yields zero subgoals, allowing us to
immediately prove any goal. *)
Theorem False_implies_nonsense :
False -> 2 + 2 = 5.
Proof.
intros contra.
inversion contra. Qed.
(** How does this work? The [inversion] tactic breaks [contra] into
each of its possible cases, and yields a subgoal for each case.
As [contra] is evidence for [False], it has _no_ possible cases,
hence, there are no possible subgoals and the proof is done. *)
(** Conversely, the only way to prove [False] is if there is already
something nonsensical or contradictory in the context: *)
Theorem nonsense_implies_False :
2 + 2 = 5 -> False.
Proof.
intros contra.
inversion contra. Qed.
(** Actually, since the proof of [False_implies_nonsense]
doesn't actually have anything to do with the specific nonsensical
thing being proved; it can easily be generalized to work for an
arbitrary [P]: *)
Theorem ex_falso_quodlibet : forall (P:Prop),
False -> P.
Proof.
(* WORKED IN CLASS *)
intros P contra.
inversion contra. Qed.
(** The Latin _ex falso quodlibet_ means, literally, "from
falsehood follows whatever you please." This theorem is also
known as the _principle of explosion_. *)
(* #################################################### *)
(** ** Truth *)
(** Since we have defined falsehood in Coq, one might wonder whether
it is possible to define truth in the same way. We can. *)
(** **** Exercise: 2 stars, optional (True_induction) *)
(** Define [True] as another inductively defined proposition. What
induction principle will Coq generate for your definition? (The
intution is that [True] should be a proposition for which it is
trivial to give evidence. Alternatively, you may find it easiest
to start with the induction principle and work backwards to the
inductive definition.) *)
(* FILL IN HERE *)
(** [] *)
(** However, unlike [False], which we'll use extensively, [True] is
just a theoretical curiosity: it is trivial (and therefore
uninteresting) to prove as a goal, and it carries no useful
information as a hypothesis. *)
(* #################################################### *)
(** * Negation *)
(** The logical complement of a proposition [P] is written [not
P] or, for shorthand, [~P]: *)
Definition not (P:Prop) := P -> False.
(** The intuition is that, if [P] is not true, then anything at
all (even [False]) follows from assuming [P]. *)
Notation "~ x" := (not x) : type_scope.
Check not.
(* ===> Prop -> Prop *)
(** It takes a little practice to get used to working with
negation in Coq. Even though you can see perfectly well why
something is true, it can be a little hard at first to get things
into the right configuration so that Coq can see it! Here are
proofs of a few familiar facts about negation to get you warmed
up. *)
Theorem not_False :
~ False.
Proof.
unfold not. intros H. inversion H. Qed.
Theorem contradiction_implies_anything : forall P Q : Prop,
(P /\ ~P) -> Q.
Proof.
(* WORKED IN CLASS *)
intros P Q H. inversion H as [HP HNA]. unfold not in HNA.
apply HNA in HP. inversion HP. Qed.
Theorem double_neg : forall P : Prop,
P -> ~~P.
Proof.
(* WORKED IN CLASS *)
intros P H. unfold not. intros G. apply G. apply H. Qed.
(** **** Exercise: 2 stars, recommended (double_neg_inf) *)
(* EXPECTED *)
(** Write an informal proof of [double_neg]:
_Theorem_: [P] implies [~~P], for any proposition [P].
_Proof_:
(* FILL IN HERE *) Admitted.
[] *)
(** **** Exercise: 2 stars, recommended (contrapositive) *)
(* EXPECTED *)
Theorem contrapositive : forall P Q : Prop,
(P -> Q) -> (~Q -> ~P).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (not_both_true_and_false) *)
(* EXPECTED *)
Theorem not_both_true_and_false : forall P : Prop,
~ (P /\ ~P).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (informal_not_PNP) *)
(** Write an informal proof (in English) of the proposition [forall P
: Prop, ~(P /\ ~P)]. *)
(* FILL IN HERE *) Admitted.
(** [] *)
Theorem five_not_even :
~ ev 5.
Proof.
(* WORKED IN CLASS *)
unfold not. intros Hev5. inversion Hev5 as [|n Hev3 Heqn].
inversion Hev3 as [|n' Hev1 Heqn']. inversion Hev1. Qed.
(** **** Exercise: 1 star (ev_not_ev_S) *)
(* EXPECTED *)
(** Theorem [five_not_even] confirms the unsurprising fact that five
is not an even number. Prove this more interesting fact: *)
Theorem ev_not_ev_S : forall n,
ev n -> ~ ev (S n).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** Note that some theorems that are true in classical logic are _not_
provable in Coq's (constructive) logic. E.g., let's look at how
this proof gets stuck... *)
Theorem classic_double_neg : forall P : Prop,
~~P -> P.
Proof.
(* WORKED IN CLASS *)
intros P H. unfold not in H.
(* But now what? There is no way to "invent" evidence for [P]. *)
Admitted.
(** **** Exercise: 5 stars, optional (classical_axioms) *)
(* NO SOLUTION *)
(** For those who like a challenge, here is an exercise
taken from the Coq'Art book (p. 123). The following five
statements are often considered as characterizations of
classical logic (as opposed to constructive logic, which is
what is "built in" to Coq). We can't prove them in Coq, but
we can consistently add any one of them as an unproven axiom
if we wish to work in classical logic. Prove that these five
propositions are equivalent. *)
Definition peirce := forall P Q: Prop,
((P->Q)->P)->P.
Definition classic := forall P:Prop,
~~P -> P.
Definition excluded_middle := forall P:Prop,
P \/ ~P.
Definition de_morgan_not_and_not := forall P Q:Prop,
~(~P/\~Q) -> P\/Q.
Definition implies_to_or := forall P Q:Prop,
(P->Q) -> (~P\/Q).
(* FILL IN HERE *)
(** [] *)
(* ########################################################## *)
(** ** Inequality *)
(** Saying [x <> y] is just the same as saying [~(x = y)]. *)
Notation "x <> y" := (~ (x = y)) : type_scope.
(** Since inequality involves a negation, it again requires
a little practice to be able to work with it fluently. Here
is one very useful trick. If you are trying to prove a goal
that is nonsensical (e.g., the goal state is [false = true]),
apply the lemma [ex_falso_quodlibet] to change the goal to
[False]. This makes it easier to use assumptions of the form
[~P] that are available in the context -- in particular,
assumptions of the form [x<>y]. *)
Theorem not_false_then_true : forall b : bool,
b <> false -> b = true.
Proof.
intros b H. destruct b.
Case "b = true". reflexivity.
Case "b = false".
unfold not in H.
apply ex_falso_quodlibet.
apply H. reflexivity. Qed.
(** **** Exercise: 2 stars, recommended (not_eq_beq_false) *)
(* EXPECTED *)
Theorem not_eq_beq_false : forall n n' : nat,
n <> n' ->
beq_nat n n' = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (beq_false_not_eq) *)
Theorem beq_false_not_eq : forall n m,
false = beq_nat n m -> n <> m.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ############################################################ *)
(** * Existential Quantification *)
(** Another critical logical connective is _existential
quantification_. We can capture what this means with the
following definition: *)
Inductive ex (X:Type) (P : X->Prop) : Prop :=
ex_intro : forall (witness:X), P witness -> ex X P.
(** That is, [ex] is a family of propositions indexed by a type [X]
and a property [P] over [X]. In order to give evidence for the
assertion "there exists an [x] for which the property [P] holds"
we must actually name a _witness_ -- a specific value [x] -- and
then give evidence for [P x], i.e., evidence that [x] has the
property [P].
For example, consider this existentially quantified proposition: *)
Definition some_nat_is_even : Prop :=
ex nat ev.
(** To prove this proposition, we need to choose a particular number
as witness -- say, 4 -- and give some evidence that that number is
even. *)
Definition snie : some_nat_is_even :=
ex_intro _ ev 4 (ev_SS 2 (ev_SS 0 ev_0)).
(** Coq's notation definition facility can be used to introduce
more familiar notation for writing existentially quantified
propositions, exactly parallel to the built-in syntax for
universally quantified propositions. Instead of writing [ex nat
ev] to express the proposition that there exists some number that
is even, for example, we can write [exists x:nat, ev x]. (It is
not necessary to understand exactly how the [Notation] definition
works.) *)
Notation "'exists' x , p" := (ex _ (fun x => p))
(at level 200, x ident, right associativity) : type_scope.
Notation "'exists' x : X , p" := (ex _ (fun x:X => p))
(at level 200, x ident, right associativity) : type_scope.
(** We can use the same set of tactics as always for
manipulating existentials. For example, if to prove an
existential, we [apply] the constructor [ex_intro]. Since the
premise of [ex_intro] involves a variable ([witness]) that does
not appear in its conclusion, we need to explicitly give its value
when we use [apply]. *)
Example exists_example_1 : exists n, n + (n * n) = 6.
Proof.
apply ex_intro with (witness:=2).
reflexivity. Qed.
(** Note, again, that we have to explicitly give the witness. *)
(** Or, instead of writing [apply ex_intro with (witness:=e)] all the
time, we can use the convenient shorthand [exists e], which means
the same thing. *)
Example exists_example_1' : exists n,
n + (n * n) = 6.
Proof.
exists 2.
reflexivity. Qed.
(** Conversely, if we have an existential hypothesis in the
context, we can eliminate it with [inversion]. Note the use
of the [as...] pattern to name the variable that Coq
introduces to name the witness value and get evidence that
the hypothesis holds for the witness. (If we don't
explicitly choose one, Coq will just call it [witness], which
makes proofs confusing.) *)
Theorem exists_example_2 : forall n,
(exists m, n = 4 + m) ->
(exists o, n = 2 + o).
Proof.
intros n H.
inversion H as [m Hm].
exists (2 + m).
apply Hm. Qed.
(** **** Exercise: 1 star, optional (english_exists) *)
(** In English, what does the proposition
ex nat (fun n => beautiful (S n))
]]
mean? *)
(* FILL IN HERE *)
(** Complete the definition of the following proof object: *)
Definition p : ex nat (fun n => beautiful (S n)) :=
(* FILL IN HERE *) admit.
(** [] *)
(** **** Exercise: 1 star (dist_not_exists) *)
(* EXPECTED *)
(** Prove that "[P] holds for all [x]" and "there is no [x] for
which [P] does not hold" are equivalent assertions. *)
Theorem dist_not_exists : forall (X:Type) (P : X -> Prop),
(forall x, P x) -> ~ (exists x, ~ P x).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (not_exists_dist) *)
(** The other direction requires the classical "law of the excluded
middle": *)
Theorem not_exists_dist :
excluded_middle ->
forall (X:Type) (P : X -> Prop),
~ (exists x, ~ P x) -> (forall x, P x).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars (dist_exists_or) *)
(* EXPECTED *)
(** Prove that existential quantification distributes over
disjunction. *)
Theorem dist_exists_or : forall (X:Type) (P Q : X -> Prop),
(exists x, P x \/ Q x) <-> (exists x, P x) \/ (exists x, Q x).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* Print dist_exists_or. *)
(* ###################################################### *)
(** * Equality *)
(** Even Coq's equality relation is not built in. It has
roughly the following inductive definition. (We enclose the
definition in a module to avoid confusion with the standard
library equality, which we have used extensively already.) *)
Module MyEquality.
Inductive eq (X:Type) : X -> X -> Prop :=
refl_equal : forall x, eq X x x.
(** Standard infix notation (using Coq's type argument synthesis): *)
Notation "x = y" := (eq _ x y)
(at level 70, no associativity) : type_scope.
(** This is a bit subtle. The way to think about it is that, given a
set [X], it defines a _family_ of propositions "[x] is equal to
[y]," indexed by pairs of values ([x] and [y]) from [X]. There is
just one way of constructing evidence for members of this family:
applying the constructor [refl_equal] to a type [X] and a value [x
: X] yields evidence that [x] is equal to [x]. *)
(** Here is a slightly different definition -- the one that actually
appears in the Coq standard library. *)
Inductive eq' (X:Type) (x:X) : X -> Prop :=
refl_equal' : eq' X x x.
Notation "x =' y" := (eq' _ x y)
(at level 70, no associativity) : type_scope.
(** **** Exercise: 3 stars, optional (two_defs_of_eq_coincide) *)
(** Verify that the two definitions of equality are equivalent. *)
Theorem two_defs_of_eq_coincide : forall (X:Type) (x y : X),
x = y <-> x =' y.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** The advantage of the second definition is that the induction
principle that Coq derives for it is precisely the familiar
principle of _Leibniz equality_: what we mean when we say "[x] and
[y] are equal" is that every property on [P] that is true of [x]
is also true of [y]. *)
Check eq'_ind.
(* ===>
forall (X : Type) (x : X) (P : X -> Prop),
P x -> forall y : X, x =' y -> P y
===> (i.e., after a little reorganization)
forall (X : Type) (x : X) forall y : X,
x =' y ->
forall P : X -> Prop, P x -> P y *)
(** One important consideration remains. Clearly, we can use
[refl_equal] to construct evidence that, for example, [2 = 2].
Can we also use it to construct evidence that [1 + 1 = 2]? Yes:
indeed, it is the very same piece of evidence! The reason is that
Coq treats as "the same" any two terms that are _convertible_
according to a simple set of computation rules. These rules,
which are similar to those used by [Eval simpl], include
evaluation of function application, inlining of definitions, and
simplification of [match]es.
In tactic-based proofs of equality, the conversion rules are
normally hidden in uses of [simpl] (either explicit or implicit in
other tactics such as [reflexivity]). But you can see them
directly at work in the following explicit proof objects: *)
Definition four : 2 + 2 = 1 + 3 :=
refl_equal nat 4.
Definition singleton : forall (X:Set) (x:X), []++[x] = x::[] :=
fun (X:Set) (x:X) => refl_equal (list X) [x].
End MyEquality.
(* ####################################################### *)
(** ** Inversion, Again *)
(** We've seen [inversion] used with both equality hypotheses and
hypotheses about inductively defined propositions. Now that we've
seen that these are actually the same thing, we're in a position
to take a closer look at how [inversion] behaves...
In general, the [inversion] tactic
- takes a hypothesis [H] whose type [P] is inductively defined,
and
- for each constructor [C] in [P]'s definition,
- generates a new subgoal in which we assume [H] was
built with [C],
- adds the arguments (premises) of [C] to the context of
the subgoal as extra hypotheses,
- matches the conclusion (result type) of [C] against the
current goal and calculates a set of equalities that must
hold in order for [C] to be applicable,
- adds these equalities to the context (and, for convenience,
rewrites them in the goal), and
- if the equalities are not satisfiable (e.g., they involve
things like [S n = O]), immediately solves the subgoal. *)
(** _Example_: If we invert a hypothesis built with [or], there are two
constructors, so two subgoals get generated. The
conclusion (result type) of the constructor ([P \/ Q]) doesn't
place any restrictions on the form of [P] or [Q], so we don't get
any extra equalities in the context of the subgoal.
_Example_: If we invert a hypothesis built with [and], there is
only one constructor, so only one subgoal gets generated. Again,
the conclusion (result type) of the constructor ([P /\ Q]) doesn't
place any restrictions on the form of [P] or [Q], so we don't get
any extra equalities in the context of the subgoal. The
constructor does have two arguments, though, and these can be seen
in the context in the subgoal.
_Example_: If we invert a hypothesis built with [eq], there is
again only one constructor, so only one subgoal gets generated.
Now, though, the form of the [refl_equal] constructor does give us
some extra information: it tells us that the two arguments to [eq]
must be the same! The [inversion] tactic adds this fact to the
context. *)
(* ####################################################### *)
(** * Relations as Propositions *)
(** A proposition parameterized by a number (such as [ev] or
[beautiful]) can be thought of as a _property_ -- i.e., it defines
a subset of [nat], namely those numbers for which the proposition
is provable. In the same way, a two-argument proposition can be
thought of as a _relation_ -- i.e., it defines a set of pairs for
which the proposition is provable. *)
Module LeFirstTry.
(** We've already seen an inductive definition of one
fundamental relation: equality. Another useful one is the "less
than or equal to" relation on numbers: *)
(** The following definition should be fairly intuitive. It
says that there are two ways to give evidence that one number is
less than or equal to another: either observe that they are the
same number, or give evidence that the first is less than or equal
to the predecessor of the second. *)
Inductive le : nat -> nat -> Prop :=
| le_n : forall n, le n n
| le_S : forall n m, (le n m) -> (le n (S m)).
End LeFirstTry.
(** This is a reasonable definition of the [<=] relation, but we
can streamline it a little by observing that the left-hand
argument [n] is the same everywhere in the definition, so we can
actually make it a "general parameter" to the whole definition,
rather than an argument to each constructor. This is similar to
what we did in our second definition of the [eq] relation,
above. *)
Inductive le (n:nat) : nat -> Prop :=
| le_n : le n n
| le_S : forall m, (le n m) -> (le n (S m)).
Notation "m <= n" := (le m n).
(** The second one is better, even though it looks less symmetric.
Why? Because it gives us a simpler induction principle.
(The same was true of our second version of [eq].) *)
Check le_ind.
(* ===> forall (n : nat) (P : nat -> Prop),
P n ->
(forall m : nat, n <= m -> P m -> P (S m)) ->
forall n0 : nat, n <= n0 -> P n0 *)
(** By contrast, the induction principle that Coq calculates for the
first definition has a lot of extra quantifiers, which makes it
messier to work with when proving things by induction. Here is
the induction principle for the first [le]: *)
(* le_ind :
forall P : nat -> nat -> Prop,
(forall n : nat, P n n) ->
(forall n m : nat, le n m -> P n m -> P n (S m)) ->
forall n n0 : nat, le n n0 -> P n n0 *)
(** Proofs of facts about [<=] using the constructors [le_n] and
[le_S] follow the same patterns as proofs about properties, like
[ev] in chapter [Prop]. We can [apply] the constructors to prove [<=]
goals (e.g., to show that [3<=3] or [3<=6]), and we can use
tactics like [inversion] to extract information from [<=]
hypotheses in the context (e.g., to prove that [~(2 <= 1)].) *)
(** Here are some sanity checks on the definition. (Notice that,
although these are the same kind of simple "unit tests" as we gave
for the testing functions we wrote in the first few lectures, we
must construct their proofs explicitly -- [simpl] and
[reflexivity] don't do the job, because the proofs aren't just a
matter of simplifying computations.) *)
Theorem test_le1 :
3 <= 3.
Proof.
(* WORKED IN CLASS *)
apply le_n. Qed.
Theorem test_le2 :
3 <= 6.
Proof.
(* WORKED IN CLASS *)
apply le_S. apply le_S. apply le_S. apply le_n. Qed.
Theorem test_le3 :
~ (2 <= 1).
Proof.
(* WORKED IN CLASS *)
intros H. inversion H. inversion H1. Qed.
(** The "strictly less than" relation [n < m] can now be defined
in terms of [le]. *)
Definition lt (n m:nat) := le (S n) m.
Notation "m < n" := (lt m n).
(** Here are a few more simple relations on numbers: *)
Inductive square_of : nat -> nat -> Prop :=
sq : forall n:nat, square_of n (n * n).
Inductive next_nat (n:nat) : nat -> Prop :=
| nn : next_nat n (S n).
Inductive next_even (n:nat) : nat -> Prop :=
| ne_1 : ev (S n) -> next_even n (S n)
| ne_2 : ev (S (S n)) -> next_even n (S (S n)).
(** **** Exercise: 2 stars, recommended (total_relation) *)
(* EXPECTED *)
(** Define an inductive binary relation [total_relation] that holds
between every pair of natural numbers. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 2 stars (empty_relation) *)
(* EXPECTED *)
(** Define an inductive binary relation [empty_relation] (on numbers)
that never holds. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars, recommended (R_provability) *)
(* EXPECTED *)
Module R.
(** We can define three-place relations, four-place relations,
etc., in just the same way as binary relations. For example,
consider the following three-place relation on numbers: *)
Inductive R : nat -> nat -> nat -> Prop :=
| c1 : R 0 0 0
| c2 : forall m n o, R m n o -> R (S m) n (S o)
| c3 : forall m n o, R m n o -> R m (S n) (S o)
| c4 : forall m n o, R (S m) (S n) (S (S o)) -> R m n o
| c5 : forall m n o, R m n o -> R n m o.
(** - Which of the following propositions are provable?
- [R 1 1 2]
- [R 2 2 6]
- If we dropped constructor [c5] from the definition of [R],
would the set of provable propositions change? Briefly (1
sentence) explain your answer.
- If we dropped constructor [c4] from the definition of [R],
would the set of provable propositions change? Briefly (1
sentence) explain your answer.
(* FILL IN HERE *) Admitted.
[] *)
(** **** Exercise: 3 stars, optional (R_fact) *)
(** State and prove an equivalent characterization of the relation
[R]. That is, if [R m n o] is true, what can we say about [m],
[n], and [o], and vice versa?
*)
(* FILL IN HERE *)
(** [] *)
End R.
(** **** Exercise: 3 stars, recommended (all_forallb) *)
(* EXPECTED *)
(** Inductively define a property [all] of lists, parameterized by a
type [X] and a property [P : X -> Prop], such that [all X P l]
asserts that [P] is true for every element of the list [l]. *)
Inductive all (X : Type) (P : X -> Prop) : list X -> Prop :=
(* FILL IN HERE *)
.
(** Recall the function [forallb], from the exercise
[forall_exists_challenge] in chapter [Poly]: *)
Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool :=
match l with
| [] => true
| x :: l' => andb (test x) (forallb test l')
end.
(** Using the property [all], write down a specification for [forallb],
and prove that it satisfies the specification. Try to make your
specification as precise as possible.
Are there any important properties of the function [forallb] which
are not captured by your specification? *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, optional (filter_challenge) *)
(* NO SOLUTION *)
(** One of the main purposes of Coq is to prove that programs match
their specifications. To this end, let's prove that our
definition of [filter] matches a specification. Here is the
specification, written out informally in English.
Suppose we have a set [X], a function [test: X->bool], and a list
[l] of type [list X]. Suppose further that [l] is an "in-order
merge" of two lists, [l1] and [l2], such that every item in [l1]
satisfies [test] and no item in [l2] satisfies test. Then [filter
test l = l1].
A list [l] is an "in-order merge" of [l1] and [l2] if it contains
all the same elements as [l1] and [l2], in the same order as [l1]
and [l2], but possibly interleaved. For example,
[1,4,6,2,3]
is an in-order merge of
[1,6,2]
and
[4,3].
Your job is to translate this specification into a Coq theorem and
prove it. (Hint: You'll need to begin by defining what it means
for one list to be a merge of two others. Do this with an
inductive relation, not a [Fixpoint].) *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars, optional (filter_challenge_2) *)
(* NO SOLUTION *)
(** A different way to formally characterize the behavior of [filter]
goes like this: Among all subsequences of [l] with the property
that [test] evaluates to [true] on all their members, [filter test
l] is the longest. Express this claim formally and prove it. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, optional (no_repeats) *)
(* EXPECTED *)
(** The following inductively defined proposition... *)
Inductive appears_in {X:Type} (a:X) : list X -> Prop :=
| ai_here : forall l, appears_in a (a::l)
| ai_later : forall b l, appears_in a l -> appears_in a (b::l).
(** ...gives us a precise way of saying that a value [a] appears at
least once as a member of a list [l].
Here's a pair of warm-ups about [appears_in].
*)
Lemma appears_in_app : forall {X:Type} (xs ys : list X) (x:X),
appears_in x (xs ++ ys) -> appears_in x xs \/ appears_in x ys.
Proof.
(* FILL IN HERE *) Admitted.
Lemma app_appears_in : forall {X:Type} (xs ys : list X) (x:X),
appears_in x xs \/ appears_in x ys -> appears_in x (xs ++ ys).
Proof.
(* FILL IN HERE *) Admitted.
(** Now use [appears_in] to define a proposition [disjoint X l1 l2],
which should be provable exactly when [l1] and [l2] are
lists (with elements of type X) that have no elements in common. *)
(* FILL IN HERE *)
(** Next, use [appears_in] to define an inductive proposition
[no_repeats X l], which should be provable exactly when [l] is a
list (with elements of type [X]) where every member is different
from every other. For example, [no_repeats nat [1,2,3,4]] and
[no_repeats bool []] should be provable, while [no_repeats nat
[1,2,1]] and [no_repeats bool [true,true]] should not be. *)
(* FILL IN HERE *)
(** Finally, state and prove one or more interesting theorems relating
[disjoint], [no_repeats] and [++] (list append). *)
(* FILL IN HERE *)
(** [] *)
(* ######################################################### *)
(** ** Digression: More Facts about [<=] and [<] *)
(** Let's pause briefly to record several facts about the [<=]
and [<] relations that we are going to need later in the
course. The proofs make good practice exercises. *)
(** **** Exercise: 2 stars, optional (le_exercises) *)
(* EXPECTED *)
Theorem O_le_n : forall n,
0 <= n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem n_le_m__Sn_le_Sm : forall n m,
n <= m -> S n <= S m.
Proof.
(* FILL IN HERE *) Admitted.
Theorem Sn_le_Sm__n_le_m : forall n m,
S n <= S m -> n <= m.
Proof.
intros n m. generalize dependent n. induction m.
(* FILL IN HERE *) Admitted.
Theorem le_plus_l : forall a b,
a <= a + b.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_lt : forall n1 n2 m,
n1 + n2 < m ->
n1 < m /\ n2 < m.
Proof.
(* FILL IN HERE *) Admitted.
Theorem lt_S : forall n m,
n < m ->
n < S m.
Proof.
(* FILL IN HERE *) Admitted.
Theorem ble_nat_true : forall n m,
ble_nat n m = true -> n <= m.
Proof.
(* FILL IN HERE *) Admitted.
Theorem ble_nat_n_Sn_false : forall n m,
ble_nat n (S m) = false ->
ble_nat n m = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem ble_nat_false : forall n m,
ble_nat n m = false -> ~(n <= m).
Proof.
(* Hint: Do the right induction! *)
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, recommended (nostutter) *)
(* EXPECTED *)
(** Formulating inductive definitions of predicates is an important
skill you'll need in this course. Try to solve this exercise
without any help at all (except from your study group partner, if
you have one).
We say that a list of numbers "stutters" if it repeats the same
number consecutively. The predicate "[nostutter mylist]" means
that [mylist] does not stutter. Formulate an inductive definition
for [nostutter]. (This is different from the [no_repeats]
predicate in the exercise above; the sequence [1,4,1] repeats but
does not stutter.) *)
Inductive nostutter: list nat -> Prop :=
(* FILL IN HERE *)
.
(** Make sure each of these tests succeeds, but you are free
to change the proof if the given one doesn't work for you.
Your definition might be different from mine and still correct,
in which case the examples might need a different proof.
The suggested proofs for the examples (in comments) use a number
of tactics we haven't talked about, to try to make them robust
with respect to different possible ways of defining [nostutter].
You should be able to just uncomment and use them as-is, but if
you prefer you can also prove each example with more basic
tactics. *)
Example test_nostutter_1: nostutter [3,1,4,1,5,6].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_false_not_eq; auto. Qed.
*)
Example test_nostutter_2: nostutter [].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_false_not_eq; auto. Qed.
*)
Example test_nostutter_3: nostutter [5].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_false_not_eq; auto. Qed.
*)
Example test_nostutter_4: not (nostutter [3,1,1,4]).
(* FILL IN HERE *) Admitted.
(*
Proof. intro.
repeat match goal with
h: nostutter _ |- _ => inversion h; clear h; subst
end.
contradiction H1; auto. Qed.
*)
(** [] *)
(** **** Exercise: 4 stars, optional (pigeonhole principle) *)
(** The "pigeonhole principle" states a basic fact about counting:
if you distribute more than [n] items into [n] pigeonholes, some
pigeonhole must contain at least two items. As is often the case,
this apparently trivial fact about numbers requires non-trivial
machinery to prove, but we now have enough... *)
(** First a pair of useful lemmas... (we already proved this for lists
of naturals, but not for arbitrary lists.) *)
Lemma app_length : forall {X:Type} (l1 l2 : list X),
length (l1 ++ l2) = length l1 + length l2.
Proof.
(* FILL IN HERE *) Admitted.
Lemma appears_in_app_split : forall {X:Type} (x:X) (l:list X),
appears_in x l ->
exists l1, exists l2, l = l1 ++ (x::l2).
Proof.
(* FILL IN HERE *) Admitted.
(** Now define a predicate [repeats] (analogous to [no_repeats] in the
exercise above), such that [repeats X l] asserts that [l] contains
at least one repeated element (of type [X]). *)
Inductive repeats {X:Type} : list X -> Prop :=
(* FILL IN HERE *)
.
(** Now here's a way to formalize the pigeonhole principle. List [l2]
represents a list of pigeonhole labels, and list [l1] represents an
assignment of items to labels: if there are more items than labels,
at least two items must have the same label. You will almost
certainly need to use the [excluded_middle] hypothesis. *)
Theorem pigeonhole_principle: forall {X:Type} (l1 l2:list X),
excluded_middle ->
(forall x, appears_in x l1 -> appears_in x l2) ->
length l2 < length l1 ->
repeats l1.
Proof. intros X l1. induction l1 as [| x l1'].
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** * Informal Proofs *)
(** Q: What is the relation between a formal proof of a proposition
[P] and an informal proof of the same proposition [P]?
A: The latter should _teach_ the reader how to produce the
former.
Q: How much detail is needed?
A: There is no single right answer; rather, there is a range
of choices.
At one end of the spectrum, we can essentially give the
reader the whole formal proof (i.e., the informal proof
amounts to just transcribing the formal one into words).
This gives the reader the _ability_ to reproduce the formal
one for themselves, but it doesn't _teach_ them anything.
At the other end of the spectrum, we can say "The theorem
is true and you can figure out why for yourself if you
think about it hard enough." This is also not a good
teaching strategy, because usually writing the proof
requires some deep insights into the thing we're proving,
and most readers will give up before they rediscover all
the same insights as we did.
In the middle is the golden mean -- a proof that includes
all of the essential insights (saving the reader the hard
part of work that we went through to find the proof in the
first place) and clear high-level suggestions for the more
routine parts to save the reader from spending too much
time reconstructing these parts (e.g., what the IH says and
what must be shown in each case of an inductive proof), but
not so much detail that the main ideas are obscured.
Another key point: if we're talking about a formal proof of a
proposition P and an informal proof of P, the proposition P doesn't
change. That is, formal and informal proofs are _talking about the
same world_ and they _must play by the same rules_. *)
(* ####################################################### *)
(** ** Informal Proofs by Induction *)
(** Since we've spent much of this chapter looking "under the hood" at
formal proofs by induction, now is a good moment to talk a little
about _informal_ proofs by induction.
In the real world of mathematical communication, written proofs
range from extremely longwinded and pedantic to extremely brief
and telegraphic. The ideal is somewhere in between, of course,
but while you are getting used to the style it is better to start
out at the pedantic end. Also, during the learning phase, it is
probably helpful to have a clear standard to compare against.
With this in mind, we offer two templates below -- one for proofs
by induction over _data_ (i.e., where the thing we're doing
induction on lives in [Type]) and one for proofs by induction over
_evidence_ (i.e., where the inductively defined thing lives in
[Prop]). In the rest of this course, please follow one of the two
for _all_ of your inductive proofs. *)
(** *** Induction Over an Inductively Defined Set *)
(** _Template_:
- _Theorem_:
_Proof_: By induction on [n].
- Suppose [n = c a1 ... ak], where <...and here we state
the IH for each of the [a]'s that has type [S], if any>.
We must show <...and here we restate [P(c a1 ... ak)]>.
- []
_Example_:
- _Theorem_: For all sets [X], lists [l : list X], and numbers
[n], if [length l = n] then [index (S n) l = None].
_Proof_: By induction on [l].
- Suppose [l = []]. We must show, for all numbers [n],
that, if length [[] = n], then [index (S n) [] =
None].
This follows immediately from the definition of index.
- Suppose [l = x :: l'] for some [x] and [l'], where
[length l' = n'] implies [index (S n') l' = None], for
any number [n']. We must show, for all [n], that, if
[length (x::l') = n] then [index (S n) (x::l') =
None].
Let [n] be a number with [length l = n]. Since
length l = length (x::l') = S (length l'),
it suffices to show that
index (S (length l')) l' = None.
]]
But this follows directly from the induction hypothesis,
picking [n'] to be length [l']. [] *)
(** *** Induction Over an Inductively Defined Proposition *)
(** Since inductively defined proof objects are often called
"derivation trees," this form of proof is also known as _induction
on derivations_.
_Template_:
- _Theorem_: P]," where [Q] is
some inductively defined proposition (more generally,
"For all [x] [y] [z], [Q x y z -> P x y z]")>
_Proof_: By induction on a derivation of [Q].
- Suppose the final rule used to show [Q] is [c]. Then
<...and here we state the types of all of the [a]'s
together with any equalities that follow from the
definition of the constructor and the IH for each of
the [a]'s that has type [Q], if there are any>. We must
show <...and here we restate [P]>.
- []
_Example_
- _Theorem_: The [<=] relation is transitive -- i.e., for all
numbers [n], [m], and [o], if [n <= m] and [m <= o], then
[n <= o].
_Proof_: By induction on a derivation of [m <= o].
- Suppose the final rule used to show [m <= o] is
[le_n]. Then [m = o] and we must show that [n <= m],
which is immediate by hypothesis.
- Suppose the final rule used to show [m <= o] is
[le_S]. Then [o = S o'] for some [o'] with [m <= o'].
We must show that [n <= S o'].
By induction hypothesis, [n <= o'].
But then, by [le_S], [n <= S o']. [] *)
(* ##################################################### *)
(** * Optional Material *)
(* ################################################### *)
(** ** Induction Principles for [/\] and [\/] *)
(** The induction principles for conjunction and disjunction are a
good illustration of Coq's way of generating simplified induction
principles for [Inductive]ly defined propositions, which we
discussed in the last chapter. You try first: *)
(** **** Exercise: 1 star, optional (and_ind_principle) *)
(** See if you can predict the induction principle for conjunction. *)
(* Check and_ind. *)
(** [] *)
(** **** Exercise: 1 star, optional (or_ind_principle) *)
(** See if you can predict the induction principle for disjunction. *)
(* Check or_ind. *)
(** [] *)
Check and_ind.
(** From the inductive definition of the proposition [and P Q]
Inductive and (P Q : Prop) : Prop :=
conj : P -> Q -> (and P Q).
we might expect Coq to generate this induction principle
and_ind_max :
forall (P Q : Prop) (P0 : P /\ Q -> Prop),
(forall (a : P) (b : Q), P0 (conj P Q a b)) ->
forall a : P /\ Q, P0 a
but actually it generates this simpler and more useful one:
and_ind :
forall P Q P0 : Prop,
(P -> Q -> P0) ->
P /\ Q -> P0
In the same way, when given the inductive definition of [or P Q]
Inductive or (P Q : Prop) : Prop :=
| or_introl : P -> or P Q
| or_intror : Q -> or P Q.
instead of the "maximal induction principle"
or_ind_max :
forall (P Q : Prop) (P0 : P \/ Q -> Prop),
(forall a : P, P0 (or_introl P Q a)) ->
(forall b : Q, P0 (or_intror P Q b)) ->
forall o : P \/ Q, P0 o
what Coq actually generates is this:
or_ind :
forall P Q P0 : Prop,
(P -> P0) ->
(Q -> P0) ->
P \/ Q -> P0
]]
*)
(* ######################################################### *)
(** ** Explicit Proof Objects for Induction *)
(** Although tactic-based proofs are normally much easier to
work with, the ability to write a proof term directly is sometimes
very handy, particularly when we want Coq to do something slightly
non-standard. *)
(** Recall the induction principle on naturals that Coq generates for
us automatically from the Inductive declation for [nat]. *)
(* Check nat_ind. *)
(* ===>
nat_ind : forall P : nat -> Prop,
P 0%nat ->
(forall n : nat, P n -> P (S n)) ->
forall n : nat, P n *)
(** There's nothing magic about this induction lemma: it's just
another Coq lemma that requires a proof. Coq generates the proof
automatically too... *)
Print nat_ind.
(* ===> (after some manual tidying)
nat_ind =
fun (P : nat -> Type)
(f : P 0)
(f0 : forall n : nat, P n -> P (S n)) =>
fix F (n : nat) : P n :=
match n as n0 return (P n0) with
| 0 => f
| S n0 => f0 n0 (F n0)
end.
*)
(** We can read this as follows:
Suppose we have evidence [f] that [P] holds on 0, and
evidence [f0] that [forall n:nat, P n -> P (S n)].
Then we can prove that [P] holds of an arbitrary nat [n] via
a recursive function [F] (here defined using the expression
form [Fix] rather than by a top-level [Fixpoint]
declaration). [F] pattern matches on [n]:
- If it finds 0, [F] uses [f] to show that [P n] holds.
- If it finds [S n0], [F] applies itself recursively on [n0]
to obtain evidence that [P n0] holds; then it applies [f0]
on that evidence to show that [P (S n)] holds.
[F] is just an ordinary recursive function that happens to
operate on evidence in [Prop] rather than on terms in [Set].
Aside to those interested in functional programming: You may
notice that the [match] in [F] requires an annotation [as n0
return (P n0)] to help Coq's typechecker realize that the two arms
of the [match] actually return the same type (namely [P n]). This
is essentially like matching over a GADT (generalized algebraic
datatype) in Haskell. In fact, [F] has a _dependent_ type: its
result type depends on its argument; GADT's can be used to
describe simple dependent types like this.
We can adapt this approach to proving [nat_ind] to help prove
_non-standard_ induction principles too. Recall our desire to
prove that
[forall n : nat, even n -> ev n].
Attempts to do this by standard induction on [n] fail, because the
induction principle only lets us proceed when we can prove that
[even n -> even (S n)] -- which is of course never provable. What
we did earlier in this chapter was a bit of a hack:
[Theorem even__ev : forall n : nat,
(even n -> ev n) /\ (even (S n) -> ev (S n))].
We can make a much better proof by defining and proving a
non-standard induction principle that goes "by twos":
*)
Definition nat_ind2 :
forall (P : nat -> Prop),
P 0 ->
P 1 ->
(forall n : nat, P n -> P (S(S n))) ->
forall n : nat , P n :=
fun P => fun P0 => fun P1 => fun PSS =>
fix f (n:nat) := match n return P n with
0 => P0
| 1 => P1
| S (S n') => PSS n' (f n')
end.
(** Once you get the hang of it, it is entirely straightforward to
give an explicit proof term for induction principles like this.
Proving this as a lemma using tactics is much less intuitive (try
it!).
The [induction ... using] tactic gives a convenient way to
specify a non-standard induction principle like this. *)
Lemma even__ev' : forall n, even n -> ev n.
Proof.
intros.
induction n as [ | |n'] using nat_ind2.
Case "even 0".
apply ev_0.
Case "even 1".
inversion H.
Case "even (S(S n'))".
apply ev_SS.
apply IHn'. unfold even. unfold even in H. simpl in H. apply H.
Qed.
(* ######################################################### *)
(** ** The Coq Trusted Computing Base *)
(** One issue that arises with any automated proof assistant is "why
trust it?": what if there is a bug in the implementation that
renders all its reasoning suspect?
While it is impossible to allay such concerns completely, the fact
that Coq is based on the Curry-Howard Correspondence gives it a
strong foundation. Because propositions are just types and proofs
are just terms, checking that an alleged proof of a proposition is
valid just amounts to _type-checking_ the term. Type checkers are
relatively small and straightforward programs, so the "trusted
computing base" for Coq -- the part of the code that we have to
believe is operating correctly -- is small too.
What must a typechecker do? Its primary job is to make sure that
in each function application the expected and actual argument
types match, that the arms of a [match] expression are constructor
patterns belonging to the inductive type being matched over and
all arms of the [match] return the same type, and so on.
There are a few additional wrinkles:
- Since Coq types can themselves be expressions, the checker must
normalize these (by using the conversion rules) before
comparing them.
- The checker must make sure that [match] expressions are
_exhaustive_. That is, there must be an arm for every possible
constructor. To see why, consider the following alleged proof
object:
Definition or_bogus : forall P Q, P \/ Q -> P :=
fun (P Q : Prop) (A : P \/ Q) =>
match A with
| or_introl H => H
end.
All the types here match correctly, but the [match] only
considers one of the possible constructors for [or]. Coq's
exhaustiveness check will reject this definition.
- The checker must make sure that each [fix] expression
terminates. It does this using a syntactic check to make sure
that each recursive call is on a subexpression of the original
argument. To see why this is essential, consider this alleged
proof:
Definition nat_false : forall (n:nat), False :=
fix f (n:nat) : False := f n.
Again, this is perfectly well-typed, but (fortunately) Coq will
reject it. *)
(** Note that the soundness of Coq depends only on the correctness of
this typechecking engine, not on the tactic machinery. If there
is a bug in a tactic implementation (and this certainly does
happen!), that tactic might construct an invalid proof term. But
when you type [Qed], Coq checks the term for validity from
scratch. Only lemmas whose proofs pass the type-checker can be
used in further proof developments. *)